What are the solutions to the equation 2x^2+12x+36=0 ?

The solutions to the equation 2x^2+12x+36=0 are x=-3+3i,x=-3-3i

Find the zeros of 2x^2+12x+36=0

The zeros of 2x^2+12x+36=0 are x=-3+3i,x=-3-3i

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2 x^{2} + 12 x + 36 = 0

x = - 3 + 3 i, x = - 3 - 3 i

Solution steps

2 x^{2} + 12 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 2 \cdot 36}{2 \cdot 2}

Simplify

1 2^{2} - 4 \cdot 2 \cdot 36 : 12 i

x_{1, 2} = \frac{- 12 \pm 12 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 12 + 12 i}{2 \cdot 2}, x_{2} = \frac{- 12 - 12 i}{2 \cdot 2}

x = \frac{- 12 + 12 i}{2 \cdot 2} : - 3 + 3 i

x = \frac{- 12 - 12 i}{2 \cdot 2} : - 3 - 3 i

The solutions to the quadratic equation are:

x = - 3 + 3 i, x = - 3 - 3 i

- 6 x^{2} + 60 x - 144 = 0

x^{2} + x + c

6^{2} = x^{2} + x^{2}

x, x^{2} + y^{2} + z^{2} = z

\frac{x ^{2}}{25} = 0.5

What are the solutions to the equation 2x^2+12x+36=0 ?
The solutions to the equation 2x^2+12x+36=0 are x=-3+3i,x=-3-3i
Find the zeros of 2x^2+12x+36=0
The zeros of 2x^2+12x+36=0 are x=-3+3i,x=-3-3i